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If $\Bbb N$ denotes all positive integers. Then find all functions $f: \mathbb{N} \to \mathbb{N}$ which are strictly increasing and such for all positive integers $n$, we have:

$$f(f(n)) = n+2$$

So far I know that $f(n)$ is greater than $n$ because the function is strictly increasing, but I'm not sure how to use this in order to solve the equation.

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  • $\begingroup$ Could we not have $f(x)=x+1$, which is strictly increasing and hence $f(f(n))=f(n+1)=n+2$? $\endgroup$ – lioness99a Jun 22 '17 at 11:41
  • $\begingroup$ hint: $ f ( n ) $ can't be greater than $ n + 2 $. $\endgroup$ – Mohsen Shahriari Jun 22 '17 at 11:42
  • $\begingroup$ strictly increasing seems to be unnecessary $\endgroup$ – Mudream Jun 29 '17 at 5:45
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    $\begingroup$ Possible duplicate of Find all functions of positive integers for $f(f(n))=n+2$ $\endgroup$ – Sil Apr 6 at 12:53
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As kindly explained below strictly increasing means $f(a) > f(b)$ for $a >b.$ Let's consider $f(0)$.

  1. $f(0) = 0$ is not possible because then $f^2(0) = 0 \neq 2.$
  2. $f(0) = 1$ is possible.
  3. $f(0) = m > 1$ is not possible, since $f^2(0) = f(m),$ and since $f$ is strictly increasing we have $f(m) > f(0) > 1$ which implies $f(m) \ge m + 2 > 2$ in contradiction with the assumption $f^2(0) = 2.$

Hence $f(0) = 1.$ Now suppose there is a $k$ such that $f(k) = k+1.$ What can you say about $f(k+1)$?

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    $\begingroup$ The identity function is strictly increasing but surely it doesn't satisfy $ f ( n ) \ge n + 1 $. $\endgroup$ – Mohsen Shahriari Jun 22 '17 at 12:23
  • $\begingroup$ you are definitely right :) Sorry for the horrible mistake. $\endgroup$ – Kore-N Jun 22 '17 at 12:35
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Hint: Set $$f(n)=an + b$$ Viz, \begin{align} f(f(n)) &= a(an+b)+b \\ &= a^2n+ b(a+1) \end{align} Thus $a= \pm 1$ and $$ b(a+1) = \begin{cases} 0, & \text{if}\ a=-1 \\ 2b, & \text{if}\ a= +1 \end{cases} $$

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    $\begingroup$ maybe other way around: 0 if a=-1, 2b if a=1? Also, the equation must be solved and b must be found? $\endgroup$ – farruhota Jun 22 '17 at 12:35
  • $\begingroup$ @FarrukhAtaev Yes! Thanks very much... $\endgroup$ – Kevin Jun 22 '17 at 13:04

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